Questions: (x+3)^2 + (y+4)^2 = 1

(x+3)^2 + (y+4)^2 = 1
Transcript text: $(x+3)^{2}+(y+4)^{2}=1$
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Solution

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Solution Steps

To express \( y \) as a function of \( x \) from the given equation of a circle, we need to solve for \( y \) in terms of \( x \). This involves isolating \( y \) on one side of the equation.

Solution Approach
  1. Start with the given equation of the circle: \((x+3)^2 + (y+4)^2 = 1\).
  2. Isolate \((y+4)^2\) by subtracting \((x+3)^2\) from both sides.
  3. Take the square root of both sides to solve for \( y + 4 \).
  4. Finally, isolate \( y \) by subtracting 4 from both sides.
Step 1: Start with the Given Equation

We start with the given equation of the circle: \[ (x+3)^2 + (y+4)^2 = 1 \]

Step 2: Isolate \((y+4)^2\)

Subtract \((x+3)^2\) from both sides: \[ (y+4)^2 = 1 - (x+3)^2 \]

Step 3: Simplify the Right-Hand Side

Simplify the expression on the right-hand side: \[ (y+4)^2 = 1 - (x+3)^2 = 1 - (x^2 + 6x + 9) = -x^2 - 6x - 8 \]

Step 4: Take the Square Root of Both Sides

Take the square root of both sides to solve for \(y + 4\): \[ y + 4 = \pm \sqrt{-(x^2 + 6x + 8)} \]

Step 5: Simplify the Square Root Expression

Simplify the expression under the square root: \[ y + 4 = \pm \sqrt{-(x+2)(x+4)} \]

Step 6: Isolate \(y\)

Subtract 4 from both sides to isolate \(y\): \[ y = -4 \pm \sqrt{-(x+2)(x+4)} \]

Final Answer

The solutions for \(y\) as a function of \(x\) are: \[ \boxed{y = -4 + \sqrt{-(x+2)(x+4)}} \quad \text{and} \quad \boxed{y = -4 - \sqrt{-(x+2)(x+4)}} \]

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